这道题我们的做法是开两个vector，分别把a和b字符的下标存进去，然后遍历a字符，我们要求长度必须大于等于k，我们可以画个图，也就是说b的下标减a的下标必须大于等于k-1 也就是b的下标必须大于等于a的下标+k-1  我们用二分找出b数组里面大于等于这个值最小的数的下标，然后用nb减去这个下标就是满足要求的个数了

#include <iostream>
#include <vector>
using namespace std;
typedef long long LL;
string s;
LL ret;
vector<LL> a, b;
char ca, cb;
LL k;
int main()
{
	cin >> k >> s >> ca >> cb;

	LL ns = s.size();
	for (int i = 0; i < ns; i++)
	{
		if (s[i] == ca) a.push_back(i);
		else if (s[i] == cb) b.push_back(i);
	}
	LL na = a.size();
	LL nb = b.size();

	for (int i = 0; i < na; i++)
	{
		LL val = a[i] + k - 1;
		LL left = 0, right = nb - 1;
		while (left < right)
		{
			LL mid = (left + right) / 2;
			if (b[mid] >= val) right = mid;
			else left = mid + 1;
		}
		if (b[left] >= val)
			ret +=  nb- left;
	}
	cout << ret << endl;
}